Traveling wave solutions of degenerate coupled multi-KdV equations

Traveling wave solutions of degenerate coupled multi-KdV equations Metin Gürses and Aslı Pekcan Citation: Journal of Mathematical Physics 57, 103507 (2016); doi: 10.1063/1.4965444 View online: http://dx.doi.org/10.1063/1.4965444 View Table of Contents: http://scitation.aip.org/content/aip/journal/jmp/57/10?ver=pdfcov Published by the AIP Publishing Articles you may be interested in Traveling wave solutions of degenerate coupled Korteweg-de Vries equation J. Math. Phys. 55, 091501 (2014); 10.1063/1.4893636 Abundant traveling wave solutions of the compound KdV-Burgers equation via the improved (G′/G)expansion method AIP Advances 2, 042163 (2012); 10.1063/1.4769751 The quasi-periodic solutions of mixed KdV equations J. Math. Phys. 53, 033508 (2012); 10.1063/1.3694731 N-soliton solutions of the KdV6 and mKdV6 equations J. Math. Phys. 51, 113507 (2010); 10.1063/1.3514121 The geometric property of soliton solutions for the integrable KdV6 equations J. Math. Phys. 51, 043508 (2010); 10.1063/1.3359002

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JOURNAL OF MATHEMATICAL PHYSICS 57, 103507 (2016)

Traveling wave solutions of degenerate coupled multi-KdV equations Metin Gürses1,a) and Aslı Pekcan2,b) 1

Department of Mathematics, Faculty of Science, Bilkent University, 06800 Ankara, Turkey Department of Mathematics, Faculty of Science, Hacettepe University, 06800 Ankara, Turkey 2

(Received 8 August 2015; accepted 4 October 2016; published online 24 October 2016) Traveling wave solutions of degenerate coupled ℓ-KdV equations are studied. Due to symmetry reduction these equations reduce to one ordinary differential equation (ODE), i.e., ( f ′)2 = Pn ( f ) where Pn ( f ) is a polynomial function of f of degree n = ℓ + 2, where ℓ ≥ 3 in this work. Here ℓ is the number of coupled fields. There is no known method to solve such ordinary differential equations when ℓ ≥ 3. For this purpose, we introduce two different types of methods to solve the reduced equation and apply these methods to degenerate three-coupled KdV equation. One of the methods uses the Chebyshev’s theorem. In this case, we find several solutions, some of which may correspond to solitary waves. The second method is a kind of factorizing the polynomial Pn ( f ) as a product of lower degree polynomials. Each part of this product is assumed to satisfy different ODEs. Published by AIP Publishing. [http://dx.doi.org/10.1063/1.4965444] I. INTRODUCTION

The system of degenerate coupled multi-field KdV equations is given as1–6 3 ut = uu x + qx2 , 2 1 2 qt = q2u x + uqx2 + qx3 , 2 .. .. .. .. . . . . 1 ℓ−1 ℓ−1 ℓ−1 qt = q u x + uqx + v x , 2 1 1 vt = − u x x x + vu x + uv x , (1) 4 2 where q1 = u and q ℓ = v. In Refs. 1–7, it was shown that this system is also a degenerate KdV system of rank one. In a previous work,8 we focused on Equation (1) for ℓ = 2. We reduced this equation into an ODE, ( f ′)2 = P4( f ), where P4( f ) is a polynomial function of degree four. We analyzed all possible cases about the zeros of P4( f ). Due to this analysis, we determined the cases when the solution is periodic or solitary. When the polynomial has one double f 2 and two simple zeros f 1 and f 3 with f 1 < f 2 < f 3, or one triple and one simple zeros, the solution is solitary. Other cases give periodic or non-real solutions. By using the Jacobi elliptic functions,9 we obtained periodic solutions and all solitary wave solutions which rapidly decay to some constants, explicitly. We have also shown that there are no real asymptotically vanishing traveling wave solutions for ℓ = 2. Indeed, we have the following theorem for the degenerate coupled ℓ-KdV equation, ℓ ≥ 2. The degenerate coupled ℓ-KdV equation can be reduced to the equation ( f ′)2 = Pℓ+2( f )

(2)

a) [email protected] b) [email protected]

0022-2488/2016/57(10)/103507/20/$30.00

57, 103507-1

Published by AIP Publishing.


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by taking ℓ functions as u(x,t) = q1(x,t) = f (ξ), q2(x,t) = f 2(ξ), . . . , v(x,t) = q ℓ = f ℓ(ξ), where ξ = x − ct in (1). Here Pℓ+2( f ) is a polynomial of f of degree ℓ + 2. If we apply the asymptotically vanishing boundary conditions to (2), we have ( f ′)2 = B f 2( f + 2c)ℓ .

(3)

Theorem 1.1. When ℓ = odd, we have B > 0, and then the degenerate coupled ℓ-KdV equation has a real traveling wave solution with asymptotically vanishing boundary conditions, but when ℓ = even, the constant B < 0. Hence the equation does not have a real traveling wave solution with asymptotically vanishing boundary conditions. Proof. Consider the degenerate coupled ℓ-KdV equation (1). Let u(x,t) = q1(x,t) = f (ξ), q (x,t) = f 2(ξ), . . . , v(x,t) = q ℓ = f ℓ(ξ), where ξ = x − ct. By using the first ℓ − 1 equations, we obtain all functions f i (ξ), i = 2, 3, . . . , ℓ as a polynomial of f (ξ). We get 2

3 2 f + d 1 = −α2 f 2 + A1( f ), 4 3 1 f 3(ξ) = Q3( f ) = c f 2 + f 3 + (c2 − d 1) f + d 2 2 2 3 = α3 f + A2( f ), ( 9 5 3 3 ) f 4(ξ) = Q4( f ) = − f 4 − c f 3 + − c2 + d 1 f 2 + (−c3 + cd 1 − d 2) f + d 3 16 2 4 4 = α4 f 4 + A3( f ), .. . f ℓ(ξ) = Q ℓ( f ) = (−1)ℓ+1α ℓ f ℓ + Aℓ−1( f ), f 2(ξ) = Q2( f ) = −c f −

where α j > 0 are constants, and Ai ( f ) and Q j ( f ) are polynomials of f of degree i and j, i = 1, 2, . . . , ℓ − 1 and j = 2, 3, . . . , ℓ, respectively. Now use the ℓth equation. We obtain ( ℓ) 1 ∂ Aℓ−1( f ) ∂ Aℓ−1( f ) 1 ′′′ f = (−1)ℓ−1α ℓ f ′ f ℓ 1 + + c(−1)ℓ−1ℓα ℓ f ′ f ℓ−1 + f ′ Aℓ−1 + f − . 4 2 2 ∂x ∂t Integrating the above equation once, we get α ℓ ℓ+1 ( ℓ) 1 ′′ f = (−1)ℓ−1 f 1+ + Rℓ( f ). 4 ℓ+1 2 By using f ′ as an integrating factor, we integrate once more. Finally, we obtain ( ℓ) 8α ℓ ( f ′)2 = (−1)ℓ−1 f ℓ+2 1 + + Rℓ+1( f ) = B f ℓ+2 + Rℓ+1( f ), (ℓ + 1)(ℓ + 2) 2

(4)

(5)

where Ri ( f ) is a polynomial of f of degree i, i = ℓ, ℓ + 1. When ℓ is odd, the coefficient of f ℓ+2, that is B, is positive, and when ℓ is even, B is negative. Applying asymptotically vanishing boundary conditions, we get ( f ′)2 = B f 2( f + 2c)ℓ, where sign(B) = (−1)ℓ−1. Hence for ℓ = odd, the degenerate coupled ℓ-KdV equation has a real traveling wave solution with asymptotically vanishing boundary conditions, but when ℓ = even, it does not. In this work, we study Equation (1) for ℓ = 3, which is 3 ut = uu x + v x , (6) 2 1 vt = vu x + uv x + ω x , (7) 2 1 1 ω t = − u x x x + ωu x + uω x , (8) 4 2 in detail. It is clear from Theorem 1.1 that unlike the case ℓ = 2, we have the real traveling wave solution with asymptotically vanishing boundary conditions in ℓ = 3 case. Reuse of AIP Publishing content is subject to the terms: https://publishing.aip.org/authors/rights-and-permissions. Downloaded to IP: 139.179.100.239 On: Mon, 24 Oct 2016 13:52:13

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Here the system (6)–(8) reduces to a polynomial of degree five f5 + 3c f 4 + (6c2 − 2d 1) f 3 + 4(c3 − cd 1 + d 2) f 2 + 8d 3 f + 8d 4 = P5( f ), (9) 2 where f (ξ) = u(x,t), and c, d 1, d 2, d 3, d 4 are constants. When the degree of the polynomial in the reduced equation is equal to five or greater, it is almost impossible to solve them. As far as we know, there is no known method to solve these equations. We shall introduce two methods to solve such equations. The first one is based on the Chebyshev’s theorem10 which is used recently to solve the Einstein field equations for a cosmological model.11,12 By using the Chebyshev’s theorem, we give several solutions of the reduced equations for ℓ = 3 and also for arbitrary ℓ. The second method is based on factorizing the polynomial Pℓ+2( f ) as the product of lower degree polynomials. In this way, we make use of the reduced equations of lower degrees. We have given all possible such solutions for ℓ = 3. The layout of our paper is as follows: In Sec. II, we study the behavior of the solutions in the neighborhood of the zeros of P5( f ) and discuss the cases giving solitary wave solutions. In Sec. III, we find exact solutions of the system (6)–(8) by a new method proposed for any ℓ ≥ 3 which uses the Chebyshev’s theorem and analyze the cases in which we may have solitary wave and kink-type solutions. In Sec. IV, we present an alternative method which is based on a factorization of polynomials. Particularly, we find the solutions of the system (6)–(8). Here we obtain many solutions for ℓ = 3 including solitary wave, kink-type, periodic, and unbounded solutions. We present some of them in the text but the rest of the solutions are given in Appendices A and B. ( f ′)2 =

II. GENERAL WAVES OF PERMANENT FORM FOR (ℓ = 3) A. Zeros of P5(f ) and types of solutions

Here we will analyze the zeros of P5( f ) in (9). (i) If f 1 = f (ξ1) is a simple zero of P5( f ), we have P5( f 1) = 0. Taylor expansion of P5( f ) about f 1 gives ( f ′)2 = P5( f 1) + P5′( f 1)( f − f 1) + O(( f − f 1)2) = P5′( f 1)( f − f 1) + O(( f − f 1)2). From here, we get f ′(ξ1) = 0 and f ′′(ξ1) = P5′( f 1)/2. Hence we can write the function f (ξ) as 1 f (ξ) = f (ξ1) + (ξ − ξ1) f ′(ξ1) + (ξ − ξ1)2 f ′′(ξ1) + O((ξ − ξ1)3) 2 1 = f 1 + (ξ − ξ1)2 P5′( f 1) + O((ξ − ξ1)3). (10) 4 Thus, in the neighborhood of ξ = ξ1, the function f (ξ) has a local minimum or maximum as P5′( f 1) is positive or negative, respectively, since f ′′(ξ1) = P5′( f 1)/2. (ii) If f 1 = f (ξ1) is a double zero of P5( f ), we have P5( f 1) = P5′( f 1) = 0. Taylor expansion of P5( f ) about f 1 gives 1 ( f ′)2 = P5( f 1) + P5′( f 1)( f − f 1) + ( f − f 1)2 P5′′( f 1) + O(( f − f 1)3) 2 1 = ( f − f 1)2 P5′′( f 1) + O(( f − f 1)3). 2 To have the real solution f , we should have P5′′( f 1) > 0. From the equality (11), we get 1  1  f ′ ± √ f P5′′( f 1) ∼ ± √ f 1 P5′′( f 1), 2 2 which gives √ ′′ ± √1 P ( f )ξ f (ξ) ∼ f 1 + αe 2 5 1 ,

(11)

(12)


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where α is a constant. Hence f → f 1 as ξ → ∓∞. The solution f can have only one peak and the wave extends from −∞ to ∞. (iii) If f 1 = f (ξ1) is a triple zero of P5( f ), we have P5( f 1) = P5′( f 1) = P5′′( f 1) = 0. Taylor expansion of P5( f ) about f 1 gives 1 1 ( f ′)2 = P5( f 1) + P5′( f 1)( f − f 1) + ( f − f 1)2 P5′′( f 1) + ( f − f 1)3 + O(( f − f 1)4) 2 6 1 3 ′′′ 4 = ( f − f 1) P5 ( f 1) + O(( f − f 1) ). 6

(13)

This is valid only if both signs of ( f − f 1)3 and P5′′′( f 1) are same. Hence, to obtain the real solution f we have the following two possibilities: (1) ( f − f 1) > 0 and P5′′′( f 1) > 0, (2) ( f − f 1) < 0 and P5′′′( f 1) < 0. If ( f − f 1) > 0 and P5′′′( f 1) > 0, then we have  1 f ′ ∼ ± √ ( f − f 1)3/2 P5′′′( f 1), 6 which gives f (ξ) ∼ f 1 + (

4 ±

√1 6



P5′′′( f 1)ξ + α1

)2 ,

(14)

where α1 is a constant. Thus f → f 1 as ξ → ±∞. Let ( f − f 1) < 0 and P5′′′( f 1) < 0. Then  1 f ′ ∼ ± √ ( f 1 − f )3/2 −P5′′′( f 1), 6 which yields 4

f (ξ) ∼ f 1 − (

±

√1 6



−P5′′′( f 1)ξ + α2

)2 ,

(15)

where α2 is a constant. Thus f → f 1 as ξ → ±∞. (iv) If f 1 = f (ξ1) is a quadruple zero of P5( f ), then we have P5( f 1) = P5′( f 1) = P5′′( f 1) = ′′′ P5 ( f 1) = 0. In this case, Taylor expansion of P5( f ) about f 1 gives ( f ′)2 =

1 ( f − f 1)4 P5(4)( f 1) + O(( f − f 1)5). 24

(16)

This is valid only if P5(4)( f 1) > 0. Then we have  1 f ′ ∼ ± √ ( f − f 1)2 P5(4)( f 1), 2 6 which gives 1

f (ξ) ∼ f 1 − ± 2√1 6



,

(17)

P5(4)( f 1)ξ + γ1

where γ1 is a constant. Thus f → f 1 as ξ → ±∞. (v) If f 1 = f (ξ1) is a zero of multiplicity 5 of P5( f ), then we have P5( f ) = ( f − f 1)5/2. This is valid only if f − f 1 > 0. So we obtain the solution f as f = f1 +

( 4 ) 1/3 9

1 (

±

ξ √ 2

+ m1

) 2/3 ,

(18)

where m1 is a constant. Hence f → f 1 as ξ → ±∞. Reuse of AIP Publishing content is subject to the terms: https://publishing.aip.org/authors/rights-and-permissions. Downloaded to IP: 139.179.100.239 On: Mon, 24 Oct 2016 13:52:13

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FIG. 1. Graphs of P5( f ) having one double and three simple zeros.

B. All possible cases giving solitary wave solutions

We analyze all possible cases about the zeros of P5( f ) that may give solitary wave solutions. Here in each cases, we will present the sketches of the graphs of P5( f ). Real solutions of ( f ′)2 = P5( f ) ≥ 0 occur in the shaded regions. 1. One double and three simple zeros

In Figure 1(b), f 1, f 3, and f 4 are simple zeros and f 2 is a double zero. The real solution occurs when f stays between f 1 and f 2 or f 2 and f 3. At f 1, P5′( f 1) = f ′′(ξ1) > 0, hence the graph of the function f is concave up at ξ1. At double zero f 2, f → f 2 as ξ → ±∞. Hence we have a solitary wave solution with amplitude f 1 − f 2 < 0. Similarly at f 3, P5′( f 3) = f ′′(ξ3) < 0, hence the graph of the function f is concave down at ξ3. Therefore, we also have a solitary wave solution with amplitude f 3 − f 2 > 0. Now consider the graph (d) in Figure 1. For f 3 we have P5′( f 3) = f ′′(ξ3) > 0, thus the graph of the function is concave up at ξ3. At double zero f 4, f → f 4 as ξ → ±∞. Hence we have a solitary wave solution with amplitude f 3 − f 4 < 0. In other cases, we have periodic solutions. 2. Two double and one simple zeros

In Figure 2(f), f 1 and f 3 are double zeros and f 2 is a simple zero. The real solution occurs when f stays between f 2 and f 3. For f 2, we have P5′( f 2) = f ′′(ξ2) > 0, thus the graph of the function is concave up at ξ2. At double zero f 3, f → f 3 as ξ → ±∞. Hence we have a solitary wave solution with amplitude f 2 − f 3 < 0. Now consider the graph (g) in Figure 2. Here f 2 and f 3 are double zeros and f 1 is a simple zero. The real solution occurs when f stays between f 1 and f 2 or f 2 and f 3. For f 1 we have P5′( f 1) = f ′′(ξ1) > 0, thus the graph of the function is concave up at ξ1. At double zero f 2, f → f 2 as ξ → ±∞. Hence we have a solitary wave solution with amplitude f 1 − f 2 < 0. The other cases give kink, anti-kink type, or unbounded solutions. 3. One triple and two simple zeros

Consider the graph (h) in Figure 3. Here f 2 and f 3 are simple zeros and f 1 is a triple zero. The real solution occurs when f stays between f 1 and f 2. For f 2 we have P5′( f 2) = f ′′(ξ2) < 0, so the graph of the function f is concave down at ξ2. At triple zero f 1, f → f 1 as ξ → ±∞. Hence we may have a solitary wave solution with amplitude f 2 − f 1 > 0. In the graph in Figure 3(i), f 1 and f 3 are simple zeros and f 2 is a triple zero. The real solution occurs when f stays between f 1 and f 2. For f 1 we have P5′( f 1) = f ′′(ξ1) > 0, so the graph of the

FIG. 2. Graphs of P5( f ) having two double and one simple zeros.


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FIG. 3. Graphs of P5( f ) having one triple and two simple zeros.

FIG. 4. Graphs of P5( f ) having one quadruple and one simple zeros.

FIG. 5. Graphs of P5( f ) having one double and one simple zeros.

function f is concave up at ξ1. At triple zero f 2, f → f 2 as ξ → ±∞. Hence we may have a solitary wave solution with amplitude f 1 − f 2 < 0. The other case gives a periodic solution. 4. One quadruple and one simple zeros

In the graph in Figure 4(l), f 1 is a simple zero and f 2 is a quadruple zero. The real solution occurs between f 1 and f 2. At f 1 we have P5′( f 1) = f ′′(ξ1) > 0, so the graph of the function f is concave up at ξ1. At quadruple zero f 2, f → f 2 as ξ → ±∞. Hence we may have a solitary wave solution with amplitude f 1 − f 2 < 0. The other case gives unbounded solution. 5. One double and one simple zeros

Consider the graph in Figure 5(n). Here f 1 is a simple zero and f 2 is a double zero. The real solution occurs between f 1 and f 2. At f 1 we have P5′( f 1) = f ′′(ξ1) > 0, so the graph of the function f is concave up at ξ1. At double zero f 2, f → f 2 as ξ → ±∞. Hence we have a solitary wave solution with amplitude f 1 − f 2 < 0. The other case gives an unbounded solution. To sum up, we can give the following proposition for ℓ = 3 case. Proposition 2.1. Equation (9) may admit solitary wave solutions when the polynomial function P5( f ) admits (i) one double and three simple zeros, (ii) two double and one simple zeros, (iii) one triple and two simple zeros, (iv) one quadruple and one simple zeros, and (v) one double and one simple zeros.

III. EXACT SOLUTIONS BY USING THE CHEBYSHEV’S THEOREM

The Chebyshev’s theorem is given as follows.10 Reuse of AIP Publishing content is subject to the terms: https://publishing.aip.org/authors/rights-and-permissions. Downloaded to IP: 139.179.100.239 On: Mon, 24 Oct 2016 13:52:13

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Theorem 3.1. Let a, b, c, α, and β be the given real numbers and α β , 0. The antiderivative,  I= x a (α + βx b )c dx, (19) is expressible by means of the elementary functions only in the three cases (1)

a+1 + c ∈ Z, b

(2)

a+1 ∈ Z, b

(3)

c ∈ Z.

(20)

The term x a (α + βx b )c is called a differential binomial. Note that the differential binomial may be expressed in terms of the incomplete beta function and the hypergeometric function. Let us define u = βx b /α. Then we have (1 + a ) a+1 1 a+1 I = α b +c β − b B y ,c − 1 . b b (a + 1 a+1 +c − a+1 1+a 1+a+b ) 1 α b β b u b F , 2 − c; ;u , (21) = 1+a b b where B y is the incomplete beta function and F(τ, κ; η; u) is the hypergeometric function. Our aim ¯ We can apply the is to transform the system (1) to ( y ′)2 = P¯ℓ+2( y) by taking f (ξ) = γ + α¯ y( βξ). ¯ Chebyshev’s theorem to this equation, if we assume that Pℓ+2( y) reduces to the form P¯ℓ+2( y) = ¯ in (9), Ay −2a (α + β y b )−2c , where −2a − 2c + b = ℓ + 2. For ℓ = 3, let u(x,t) = f (ξ) = γ + α¯ y( βξ) then the equation becomes ( y ′)2 = P¯5( y) = α1 y 5 + α2 y 4 + α3 y 3 + α4 y 2 + α5 y + α6,

(22)

where α1 = α2 = α3 = α4 = α5 = α6 =

α¯ 3 , 2 β¯ 2 ) 1 ( 5α¯ 2γ 2 + 3 α ¯ c , β¯ 2 2 ) 1 ( 5αγ ¯ 2 + 6αc ¯ 2 − 2αd ¯ 1 + 12αcγ ¯ , 2 β¯ ) 1 ( − 4cd 1 + 4d 2 + 18cγ 2 − 6d 1γ + 5γ 3 + 4c3 + 18c2γ , 2 ¯ β ) 1 ( 5γ 4 8d 3 + 12cγ 3 + + 18c2γ 2 − 8cd 1γ + 8c3γ + 8d 2γ − 6d 1γ 2 , 2 2 α¯ β¯ ( ) 1 γ5 2 3 2 3 3 2 2 8d γ + 6c γ + + 8d − 4cd γ − 2d γ + 4c γ + 4d γ . 3 4 1 1 2 2 α¯ 2 β¯ 2

To apply the Chebyshev’s theorem 3.1 we assume that P¯5( y) reduces to the following form: P¯5( y) = C y −2a (α + β y b )−2c ,

(23)

where a, b, and c are the constants in Theorem 3.1, b − 2a − 2c = 5, and C is a constant. Here we present the cases mentioned in Proposition 2.1. Other cases are given in Appendix A. (1) Let P¯5( y) = y(α + β y 2)2. This form corresponds to the case of one simple or one simple and two double zeros. We have y ′ = ± y 1/2(α + β y 2), so  y −1/2(α + β y 2)−1d y = ±ξ + A. (24) Here a = −1/2, b = 2, and c = −1. Hence (1)

a+1 + c = −3/4 < Z, b

(2)

a+1 = 1/4 < Z, b

(3)

c = −1 ∈ Z.


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FIG. 6. Graph of a solution of (y ′)2 = y(−1 + y 2)2.

For (3), from (24), by letting u = β y b /α, we obtain 5 ) , 3; ; u = ±ξ + A. (25) 4 4 Here choosing α = −1, β = 1, the integration constant A = 2, and the plus sign in (24), we have √ √ (26) arctanh( y) + arctan( y) = −(ξ + 2) 2α −3/4 β −1/4u1/4F

(1

and the graph of this solution for ξ ≤ −2 is given in Figure 6. This is a kink-type solution. (2) Let P¯5( y) = y(α + β y)4. This form corresponds to the case of one simple and one quadruple zeros. We have y ′ = ± y 1/2(α + β y)2, so  y −1/2(α + β y)−2dy = ±ξ + A. (27) Here a = −1/2, b = 1, and c = −2. Hence (1)

a+1 + c = −3/2 < Z, b

(2)

a+1 = 1/2 < Z, b

(3)

c = −2 ∈ Z.

For (3), from (27), by letting u = β y b /α, we obtain ( β√ y ) √ arctan √ βα ) y 3 = ±ξ + A. + 2α −3/2 β −1/2u1/2F , 4; ; u = √ 2 2 α( β y + α) α βα (1

(28)

Here choosing α = 1, β = 1, the integration constant A = 2, and the plus sign in (27), we have √ y √ + arctan( y) = ξ + 2 (29) 1+ y and the graph of this solution for −2 ≤ ξ < π/2 − 2 is given in Figure 7.

FIG. 7. Graph of a solution of (y ′)2 = y(1 + y)4.


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(3) Let P¯5( y) = y 2(α + β y 3). This form corresponds to the case of one simple and one double zeros. We have y ′ = ± y(α + β y 3)1/2, so  y −1(α + β y 3)−1/2d y = ±ξ + A. (30) Here a = −1, b = 3, and c = −1/2. Hence a+1 + c = −1/2 < Z, b For (2), from (30), we obtain (1)

y=

(α ( β

(2)

tanh2

a+1 = 0 ∈ Z, b

(3√ 2

(3)

c = −1/2 < Z.

) ) ) 1/3 α(A ± ξ) − 1 .

(31)

Here choosing α = 4, β = 4, the integration constant A = 2, and the plus sign in (30), we obtain y(ξ) = −(sech(3ξ + 6))2/3

(32)

and the graph of this solution is given in Figure 8. This is clearly a solitary wave solution. (4) Let P¯5( y) = y 2(α + β y)3. This form corresponds to the case of one double and one triple zeros. We have y ′ = ± y(α + β y)3/2, so  y −1(α + β y)−3/2d y = ±ξ + A. (33) Here a = −1, b = 1, and c = −3/2. Hence a+1 + c = −3/2 < Z, b For (2), from (33), we obtain (1)

(2)

a+1 = 0 ∈ Z, b

(3)

c = −3/2 < Z.

) (√ 2arctanh β√y+α 2 α = ±ξ + A. − √ α 3/2 α βy + α

(34)

Here choosing α = 1, β = 1, A = 2, and the plus sign in (33), we have 2 

y +1

 − 2arctanh( y + 1) = ξ + 2

(35)

and the graph of this solution is given in Figure 9. (5) Let P¯5( y) = y 4(α + β y). This form corresponds to the case of one simple and one quadruple zeros. We have y ′ = ± y 2(α + β y)1/2, so  y −2(α + β y)−1/2d y = ±ξ + A. (36)

FIG. 8. Graph of a solution of (y ′)2 = y 2(4 + 4y 3).


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FIG. 9. Graph of a solution of (y ′)2 = y 2(1 + y)3.

Here a = −2, b = 1, and c = −1/2. Hence (1)

a+1 + c = −3/2 < Z, b

(2)

a+1 = −1 ∈ Z, b

c = −1/2 < Z.

(3)

For (2), from (36), by letting u = β y b /α, we obtain 5 − α −3/2 βu−1F − 1, ; 0; u = − 2 (

)

√

βy + α + αy

βarctanh

( √ β y+α ) √

α 3/2

α

= ±ξ + A.

(37)

Taking α = 1, β = 1, A = 2, and the plus sign in (36), we get   arctanh( y + 1) −

y +1 = ξ +2 y

(38)

and the graph of this solution for ξ ≥ −2 is given in Figure 10.

FIG. 10. Graph of a solution of (y ′)2 = y 4(1 + y).


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IV. AN ALTERNATIVE METHOD TO SOLVE (f ′)2 = Pn(f ), WHEN n ≥ 5: FACTORIZATION OF POLYNOMIALS

When we have ( f ′)2 = Pn ( f ) for the reduced equation, it becomes quite difficult to solve such equations for n ≥ 5. For this purpose, we shall introduce a new method which is based on the factorization of the polynomial Pn ( f ) = Pℓ+2( f ), n ≥ 5. Let the polynomial Pℓ+2( f ) have ℓ + 2 real ℓ+2  roots, i.e., Pℓ+2( f ) = B ( f − f i ), B is a constant. Define a new function ρ(ξ), so that f = f (ρ(ξ)). i=1 ( df ) 2 ( dρ ) 2 Hence we have ( f ′)2 = dρ dξ . By taking ( d f )2 dρ

=κ

N1  ( f − f i ), i=1

N2  ( dρ ) 2 =µ ( f − f i ), dξ i=1

where N1 + N2 = ℓ + 2 and B = κ µ, we get a system of ordinary differential equations. Solving this system gives the solution of the degenerate coupled ℓ-KdV equation. For illustration, we start with a differential equation where we know the solution. Consider ( y ′)2 = α( y − y1)( y − y2)( y − y3)( y − y4).

(39)

Let y = y(ρ(ξ)) so ( d y ) 2 ( dρ ) 2 dρ

dξ

= α( y − y1)( y − y2)( y − y3)( y − y4).

Taking ( dy )2

= κ( y − y1)( y − y2), dρ ( dρ ) 2 = µ( y − y3)( y − y4), dξ

(40) (41)

where κ µ = α. We start with solving Equation (40). We have √ dy = ± κdρ, = ( ) 2 y−a ( y − y1)( y − y2) b −1 b dy



√ where a = ( y1 + y2)/2 and b = ( y1 − y2)/2. Let ( y − a)/b = cosh z. This gives us dz = ± κdρ. After taking the integral, we obtain ( y − a) √ arccosh = ± κ ρ + C1, C1 constant, b so that √ y = h(ρ) = a + b cosh( κ ρ + C2), C2 constant. (42) √ Now we insert this solution into Eq. (41). Let cosh( κ ρ + C2) = v. Hence we get  dv √ = ± µξ + C3, (43)  √ √ 2 κb v − 1 (v − A)2 − B 2 where A = ( y3 + y4 − y1 − y2)/( y1 − y2), B = ( y3 − y4)/( y1 − y2), and C3 is a constant. Solving (43) and using (42) gives  √ y1( y3 − y2) + y2( y1 − y3)sn2((1/2) κ( y3 − y2)( y4 − y1)( µξ + C3), k) y= , (44)  √ ( y3 − y2) + ( y1 − y3)sn2((1/2) κ( y3 − y2)( y4 − y1)( µξ + C3), k) which was obtained in Ref. 8. Here sn is the Jacobi elliptic function and k is the elliptic modulus satisfying k 2 = ( f 2 − f 4)( f 1 − f 3)/( f 2 − f 3)( f 1 − f 4). Reuse of AIP Publishing content is subject to the terms: https://publishing.aip.org/authors/rights-and-permissions. Downloaded to IP: 139.179.100.239 On: Mon, 24 Oct 2016 13:52:13

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Now we apply our method to the case when the polynomial Pn ( f ) is of degree n = 5. We have several possible cases, but here we shall give the case when the polynomial (9) has five real roots. Other cases are presented in Appendix B. Case 1. If (9) has five real roots, we can write it in the form ( f ′)2 = α( f − f 1)( f − f 2)( f − f 3)( f − f 4)( f − f 5),

(45)

where f 1, f 2, f 3, f 4, and f 5 are the zeros of the polynomial function P5( f ). Now define a new function ρ(ξ), so that f = f (ρ(ξ)). Hence (45) becomes ( f ′)2 =

( d f ) 2 ( dρ ) 2 dρ

dξ

= α( f − f 1)( f − f 2)( f − f 3)( f − f 4)( f − f 5).

(46)

(i) Take ( d f )2

= −( f − f 1)( f − f 2)( f − f 3)( f − f 4), dρ ( dρ ) 2 = −α( f − f 5). dξ In Ref. 8, we have found the solutions of Eq. (47). One of the solutions is  f 4( f 3 − f 1) + f 3( f 1 − f 4)sn2((1/2) ( f 1 − f 3)( f 2 − f 4)ρ, k) f = h(ρ) = ,  ( f 3 − f 1) + ( f 1 − f 4)sn2((1/2) ( f 1 − f 3)( f 2 − f 4)ρ, k)

(47) (48)

(49)

where k is the elliptic modulus satisfying k 2 = ( f 3 − f 2)( f 4 − f 1)/( f 3 − f 1)( f 4 − f 2). We use this solution in Equation (48) and get  ρ  ρ √ A + Bsn2(ω ρ, ˆ k) d ρˆ = d ρˆ = ± αξ + C1, C1 constant, (50)  2 C + Dsn (ω ρ, ˆ k) 0 0 f 5 − h( ρ) ˆ where A = f 3 − f 1, B = f 1 − f 4, C = ( f 5 − f 4)( f 3 − f 1), D = ( f 5 − f 3)( f 1 − f 4),  and ω = (1/2) ( f 1 − f 3)( f 2 − f 4). For particular values, f 1 = f 2 = −1, f 3 = 3, f 4 = 0, f 5 = 2, α = 1, C1 = 0, and choosing the plus sign in (50), we get the graph of the solution f (ξ) given in Figure 11.

FIG. 11. Graph of the solution (49) with (50) for particular parameters.


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This is a solitary wave solution. (ii) Take ( d f )2

= κ( f − f 1)( f − f 2)( f − f 3), dρ ( dρ ) 2 = µ( f − f 4)( f − f 5), dξ

(51) (52)

where κ µ = α. Eq. (51) has a solution  √ f = h(ρ) = ( f 2 − f 1)sn2((1/2) f 3 − f 1( κ ρ + C1), k) + f 1,

(53)

where k satisfies k 2 = ( f 1 − f 2)/( f 1 − f 3) and C1 is a constant. We use this solution in Eq. (52) and get  ρ  ρ d ρˆ d ρˆ √ = = ± µξ + C2, (54)   4 2 0 0 (h( ρ) ˆ − f 4)(h( ρ) ˆ − f 5) Asn (ω ρˆ + δ, k) + Bsn (ω ρˆ + δ, k) + C where C2 is a constant and  A = ( f 2 − f 1)2, B = ( f 2 − f 1)(2 f 1 − f 4 − f 5), C = ( f 1 − f 4)( f 1 − f 5), ω = (1/2) κ( f 3 − f 1),  and δ = (1/2) f 3 − f 1C1. For particular values, f 1 = 1, f 2 = 2, f 3 = 3, f 4 = −1, f 5 = −2, κ = µ = 1, C1 = 0, and choosing the plus sign in (54), we get the graph of the solution f (ξ) given in Figure 12. This solution is periodic. (iii) Take ( d f )2

= κ( f − f 1)( f − f 2), dρ ( dρ ) 2 = µ( f − f 3)( f − f 4)( f − f 5), dξ

(55) (56)

where κ µ = α. Consider first Equation (55). We have df 

( f − f 1)( f − f 2)

√ = ± κdρ.

(57)

FIG. 12. Graph of the solution (53) with (54) for particular parameters.


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This equality can be reduced to √ df = ± κdρ, ( ) f −a 2 b −1 b

(58)

where a = ( f 1 + f 2)/2 and b = ( f 1 − f 2)/2. By making the change of variables ( f − a)/b = u, we get  √ du = ± κ ρ + C1, (59) √ u2 − 1 where C1 is a constant. Thus, we obtain ( f − a) √ (60) arccoshu = arccosh = ± κ ρ + C1, b which yields √ f = h(ρ) = a + b cosh( κ ρ + C1), a = ( f 1 + f 2)/2, b = ( f 1 − f 2)/2. (61) Now we use this result in (56),  ρ d ρˆ =  0 (h( ρ) ˆ − f 3)(h( ρ) ˆ − f 4)(h( ρ) ˆ − f 5)  ρ d ρˆ =  √ √ √ 0 A cosh3( κ ρˆ + C1) + B cosh2( κ ρˆ + C1) + C cosh( κ ρˆ + C1) + D √ = ± µξ + C2,

(62)

where C2 is a constant and A = b3, B = b2(3a − f 3 − f 4 − f 5), C = b{(a − f 3)(a − f 4) + (a − f 3)(a − f 5) + (a − f 4)(a − f 5)}, and D = (a − f 3)(a − f 4)(a − f 5). For particular values, f 1 = 5, f 2 = 1, f 3 = f 4 = −3, f 5 = −4, κ = 4, µ = 1, C1 = C2 = 0, and choosing the plus sign in (62), we get the graph of the solution f (ξ) given in Figure 13. (iv) Take ( d f )2 = κ( f − f 1), (63) dρ ( dρ ) 2 = µ( f − f 2)( f − f 3)( f − f 4)( f − f 5), (64) dξ

FIG. 13. Graph of the solution (61) for particular parameters.


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FIG. 14. Graph of the solution (66) for particular parameters.

where κ µ = α. Consider Eq. (63). We have df 

f − f1

√ = ± κdρ.

(65)

Integrating both sides gives √ (

f = h(ρ) = ±

)2 κ ρ + C1 + f 1, 2

C1

constant.

(66)

We use this result in Equation (64),  ρ d ρˆ =  0 (h( ρ) ˆ − f 2)(h( ρ) ˆ − f 3)(h( ρ) ˆ − f 4)(h( ρ) ˆ − f 5)  ρ d ρˆ =  √ √ √ √ κ 2 κ 2 0 ( 4 ρˆ ± κC1 ρˆ + A)( 4 ρˆ ± κC1 ρˆ + B)( κ4 ρˆ2 ± κC1 ρˆ + C)( κ4 ρˆ2 ± κC1 ρˆ + D) √ (67) = ± µξ + C2, where C2 is a constant and A = C12 + f 1 − f 2, B = C12 + f 1 − f 3, C = C12 + f 1 − f 4, D = C12 + f 1 − f 5. For particular values, f 1 = 1, f 2 = f 3 = 2, f 4 = 3, f 5 = 4, κ = 4, µ = 1, C1 = C2 = 0, and choosing the plus sign in (67), we get the graph of the solution f (ξ) given in Figure 14. This is a solitary wave solution.

V. CONCLUSION

We study the degenerate ℓ-coupled KdV equations. We reduce these equations into an ODE of the form ( f ′)2 = Pℓ+2( f ), where Pℓ+2 is a polynomial function of f of degree ℓ + 2, ℓ ≥ 3. We give a general approach to solve the degenerate ℓ-coupled equations by introducing two new methods that one of them uses the Chebyshev’s theorem and the other one is an alternative method, based on the factorization of Pℓ+2( f ), ℓ ≥ 3. Particularly, for the degenerate three-coupled KdV equations, we obtain solitary-wave, kink-type, periodic, or unbounded solutions by using these methods. ACKNOWLEDGMENTS

This work is partially supported by the Scientific and Technological Research Council of Turkey (TÜB˙ITAK). Reuse of AIP Publishing content is subject to the terms: https://publishing.aip.org/authors/rights-and-permissions. Downloaded to IP: 139.179.100.239 On: Mon, 24 Oct 2016 13:52:13

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APPENDIX A: OTHER CASES USING CHEBYSHEV’S THEOREM FOR DEGENERATE THREE-COUPLED KdV EQUATION

Here we present other cases that we use the Chebyshev’s theorem to solve degenerate threecoupled KdV equation. (1) Let P¯5( y) = y(α + β y 4). This form corresponds to the case of one simple zero or three simple zeros. We have y ′ = ± y 1/2(α + β y 4)1/2, so  y −1/2(α + β y 4)−1/2d y = ±ξ + A. Here a = −1/2, b = 4, and c = −1/2. So we have a+1 a+1 + c = −3/8 < Z, (2) = 1/8 < Z, (3) c = −1/2 < Z. b b Hence we cannot obtain a solution through the Chebyshev’s theorem. (2) Let P¯5( y) = y 3(α + β y 2). This form corresponds to the case of one triple or two simple and one triple zeros. We have y ′ = ± y 3/2(α + β y 2)1/2, so  y −3/2(α + β y 2)−1/2d y = ±ξ + A. (1)

Here a = −3/2, b = 2, and c = −1/2. So we have a+1 a+1 + c = −3/4 < Z, (2) = −1/4 < Z, (3) c = −1/2 < Z. b b Hence we cannot obtain a solution through the Chebyshev’s theorem. (3) Let P¯5( y) = y 3(α + β y)2. This form corresponds to the case of one double and one triple zeros. We have y ′ = ± y 3/2(α + β y), so  y −3/2(α + β y)−1d y = ±ξ + A. (A1) (1)

Here a = −3/2, b = 1, and c = −1. Hence (1)

a+1 + c = −3/2 < Z, b

(2)

a+1 = −1/2 < Z, b

(3)

c = −1 ∈ Z.

For (3), from (A1), by letting u = β y b /α, we obtain (√ ) √ 2 β arctan √βαy ) 1 1 2 − 2α −3/2 β 1/2u−1/2F − , 3; ; u = − √ + = ±ξ + A. 2 2 α y α 3/2 (

(A2)

(4) Let P¯5( y) = (α + β y)5. This form corresponds to the case of a zero with multiplicity five. We have y ′ = ±(α + β y)5/2, so  (α + β y)−5/2dy = ±ξ + A. (A3) Here a = 0, b = 1, and c = −5/2. Hence (1)

a+1 + c = −3/2 < Z, b

(2)

a+1 = 1 ∈ Z, b

(3)

c = −5/2 < Z.

For (2), from (A3), by letting u = β y b /α, we obtain ( 9 ) 2 α −3/2 β −1uF 1, ; 2; u = − = ±ξ + A. 2 3 β( β y + α)3/2

(A4)

Here we get y=

)  1 ( 4 − α . β 9 β 2(±ξ + A)2

(A5)


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(5) Let P¯5( y) = (α + β y 5). This form corresponds to the case of one simple zero. We have y ′ = ±(α + β y 5)1/2, so  (α + β y 5)−1/2d y = ±ξ + A. Here a = 0, b = 5, and c = −1/2. We have a+1 a+1 + c = −3/10 < Z, (2) = 1/5 < Z, (3) c = −1/2 < Z. b b Hence we cannot obtain a solution through the Chebyshev’s theorem. (1)

APPENDIX B: OTHER CASES USING ALTERNATIVE METHOD FOR DEGENERATE THREE-COUPLED KdV EQUATION

Here we present the other cases for the method of factorization of the polynomial P5( f ). Case 2. If (9) has three real roots, we can write it in the form ( f ′)2 = α( f − f 1)( f − f 2)( f − f 3)(a f 2 + b f + c),

(B1)

where f 1, f 2, and f 3 are the zeros of the polynomial function P5( f ) of degree five and b2 − 4ac < 0. Now define a new function ρ(ξ), so that f = f (ρ(ξ)). Hence (B1) becomes ( f ′)2 =

( d f ) 2 ( dρ ) 2 dρ

dξ

= α( f − f 1)( f − f 2)( f − f 3)(a f 2 + b f + c).

(B2)

(i) Take ( d f )2

= κ( f − f 1)( f − f 2)( f − f 3), dρ ( dρ ) 2 = µ(a f 2 + b f + c), dξ

(B3) (B4)

where κ µ = α. Consider the first equation above. We have df 

( f − f 1)( f − f 2)( f − f 3)

√ = ± κdρ.

(B5)

We know from Case 1(ii) that Eq. (B3) has a solution  √ f = h(ρ) = ( f 2 − f 1)sn2((1/2) f 3 − f 1( κ ρ + C1)) + f 1,

(B6)

where C1 is a constant. We use this solution in (B4), and we get  ρ  ρ d ρˆ d ρˆ √ = = ± µξ + C2, (B7)   2 4 2 0 0 ah ( ρ) ˆ + bh( ρ) ˆ +c Asn (ω ρˆ + δ) + Bsn (ω ρˆ + δ) + C where C2 is a constant and  A = a( f 2 − f 1)2, B = (2a f 1 + b)( f 2 − f 1), C = (a f 12 + b f 1 + c), ω = (1/2) κ( f 3 − f 1),  and δ = (1/2) f 3 − f 1C1. (ii) Take ( d f )2

= κ( f − f 1)( f − f 2), dρ ( dρ ) 2 = µ( f − f 3)(a f 2 + b f + c), dξ where κ µ = α. Consider first Equation (B8). It has a solution √ f = h(ρ) = a1 + b1 cosh( κ ρ + C1), a1 = ( f 1 + f 2)/2,

(B8) (B9)

b1 = ( f 1 − f 2)/2,

(B10)


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as it is found in Case 1(iii). We use this solution in (B9) and get  ρ d ρˆ =  0 (h( ρ) ˆ − f 3)(ah2( ρ) ˆ + bh( ρ) ˆ + c)  ρ d ρˆ =  √ √ √ 0 A cosh3( κ ρˆ + C1) + B cosh2( κ ρˆ + C1) + C cosh( κ ρˆ + C1) + D √ = ± µξ + C2,

(B11)

where C2 is a constant and A = ab31, B = b21(3aa1 + b − f 3), C = b1(3aa12 − 2aa1 f 3 + 2ba1 − f 3b + c), and D = (a1 − f 3)(a12a + ba1 + c). (iii) Take ( d f )2

= κ( f − f 1), (B12) dρ ( dρ ) 2 = µ( f − f 2)( f − f 3)(a f 2 + b f + c), (B13) dξ where κ µ = α. Consider Equation (B12). It has a solution )2 ( √κ ρ + C1 + f 1, C1 constant, (B14) f = h(ρ) = ± 2 as it is obtained in Case 1(iv). We use this result in Eq. (B13)  ρ d ρˆ =  0 (h( ρ) ˆ − f 2)(h( ρ) ˆ − f 3)(ah2( ρ) ˆ + bh( ρ) ˆ + c)  ρ d ρˆ =  √ √ 3/2 2 0 ˆ4 ± aκ2 C1 ρˆ3 + C ρˆ2 + D ρˆ + E) ( κ4 ρˆ2 ± κC1 ρˆ + A)( κ4 ρˆ2 ± κC1 ρˆ + B)( aκ 16 ρ √ = ± µξ + C2, (B15) where C2 is a constant and A = C12 + f 1 − f 2, B = C12 + f 1 − f 3, C = κ(b + 6aC12 + 2a f 1)/4 and √ D = ± κC1(2a f 1 + 2aC12 + b), E = ( f 1 + C12)(a f 1 + aC12 + b) + c. (iv) Take ( d f )2

= κ(a f 2 + b f + c), dρ ( dρ ) 2 = µ( f − f 1)( f − f 2)( f − f 3), dξ where κ µ = α. Consider the first equation above. We have  f √ d fˆ = ± κ ρ + C1,  ( ) b 2 0 √ fˆ+ 2a a +1 W ( ) √ b where W = 4ac − b2/2a. Let f + 2a /W = tan θ. Hence the above equality becomes  θ √ √ ˆ θˆ = a(± κ ρ + C1), sec θd

(B16) (B17)

0

which gives ln | sec θ + tan θ| =

√

√ a(± κ ρ + C1).


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Hence after some simplifications and taking the integration constant zero, we get √ √ f = h(ρ) = ± 4ac − b2 sinh( aκ ρ)/2a. We use this result in (B17),  ρ  0

 =

0

ρ

(B18)

d ρˆ

= (h( ρ) ˆ − f 1)(h( ρ) ˆ − f 2)(h( ρ) ˆ − f 3) d ρˆ  √ √ √ A sinh3( aκ ρ) ˆ + B sinh2( aκ ρ) ˆ + C sinh( aκ ρ) ˆ +D

√ = ± µξ + C2,

(B19)

where C2 is a constant and A = ±(4ac − b2)−3/2/8a3, B = (b2 − 4ac)( f 1 + f 2 + f 3)/4a2, and

√ C = ± 4ac − b2( f 1 f 3 + f 2 f 3 + f 1 f 2)/2a, D = − f 1 f 2 f 3. Case 3. If (9) has just one real root, we can write it in the form ( f ′)2 = α( f − f 1)( f 4 + a3 f 3 + a2 f 2 + a1 f + a0),

(B20)

where f 1 is the zero of the polynomial function P5( f ) of degree five and the constants ai , i = 0, 1, 2, 3 are so that f 4 + a3 f 3 + a2 f 2 + a1 f + a0 , 0 for real f . Now define a new function ρ(ξ), so that f = f (ρ(ξ)). Hence (B20) becomes ( d f ) 2 ( dρ ) 2 ( f ′)2 = = α( f − f 1)( f 4 + a3 f 3 + a2 f 2 + a1 f + a0). (B21) dρ dξ (i) Take ( d f )2 = κ( f − f 1), (B22) dρ ( dρ ) 2 = µ( f 4 + a3 f 3 + a2 f 2 + a1 f + a0), (B23) dξ where κ µ = α. Consider the first equation above. It has a solution ( √κ )2 f = h(ρ) = ± ρ + C1 + f 1, C1 constant, (B24) 2 as it is obtained in Case 1(iv). We use this result in (B23) and get  ρ d ρˆ √ = ± µξ + C2, (B25)  4 3 2 0 h ( ρ) ˆ + a3 h ( ρ) ˆ + a2 h ( ρ) ˆ + a1 h( ρ) ˆ + a0 where C2 is a constant. Here we do not get a simpler expression than the original equation. Hence it is meaningless to use the method of factorization of the polynomial for this case. (ii) Take ( d f )2 = κ( f 4 + a3 f 3 + a2 f 2 + a1 f + a0), (B26) dρ ( dρ ) 2 = µ( f − f 1), (B27) dξ where κ µ = α. Here for simplicity, we will use the form f 4 + b f 2 + c instead of the polynomial function of degree four above. Since f 4 + b f 2 + c = ( f 2 + b/2)2 − b2/4 + c, we will take c > b2/4 ( df ) 2 to get an irreducible polynomial. If c = b2/4, then take b > 0. From dρ = κ( f 4 + b f 2 + c),  f √ d fˆ = ± κ ρ + C1, C1 constant.  0 fˆ4 + b fˆ2 + c Reuse of AIP Publishing content is subject to the terms: https://publishing.aip.org/authors/rights-and-permissions. Downloaded to IP: 139.179.100.239 On: Mon, 24 Oct 2016 13:52:13

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We have

J. Math. Phys. 57, 103507 (2016)

√

where k =

 √ √ √ 2c f (ρ) =  sn(( 2/2)(C1 ± κ ρ) −b + −4c + b2, k), √ −b + −4c + b2  √

√

−2c+b 2+b −4c+b 2 . c

2 2

 0

Then we use this function f in (B27) and get

√

ρ

√

Ad ρˆ √ = ± µξ + C2, √ √ 2c sn((A 2/2)(C1 ± κ ρ), k) − A f 1

 √ where A = −b + −4c + b2. Now take c = b2/4 with b > 0. In this case, we have  f √ d fˆ = ± κ ρ + C1, 2 ˆ 0 f + b/2

C1

C2 constant,

(B28)

constant,

which gives the solution √

√ ( 2b √ ) 2b f = h(ρ) = ± tan ( κ ρ + C˜1) . 2 2 Now use Equation (B27). We have  ρ d ρˆ √ = ± µξ + C2,  √ (√ √ ) 0 ± 22b tan 22b ( κ ρˆ + C˜1) − f 1

(B29)

where C2 is a constant. Solving the above equation gives √  2b  (A1 − A2)2 + A23  ( 2A1 A3 )  ±4 √ 2 A ln − b arctan = ± µξ + C2, √ 3 2 2 2 2 2 2 2 4 (A1 + A2) + A3 A3 − A1 + A2 κb(A2 + A3)A3 where  A1 = and A3 =



 √ ( 2b √ ) ˜ ±2 2b tan 4 f 12 + 2b − 2 f 1, ( κ ρ + C1) − 4 f 1, A2 = 2 √

4 f 12 + 2b + 2 f 1.

1

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